Partitioning two-coloured complete multipartite graphs ... - DIM-UChile

Partitioning two-coloured complete multipartite graphs into monochromatic paths and cycles Oliver Schaudt Institut f¨ ur Informatik, Universit¨at zu K¨oln, K¨oln, Germany [email protected] Maya Stein∗ Centro de Modelamiento Matem´atico, Universidad de Chile, Santiago, Chile [email protected] October 2, 2014

Abstract We show that any complete k-partite graph G on n vertices, with k ≥ 3, whose edges are two-coloured, can be covered with two vertex-disjoint monochromatic paths of distinct colours. We prove this under the necessary assumption that the largest partition class of G contains at most n/2 vertices. This extends known results for complete and complete bipartite graphs. Secondly, we show that in the same situation, all but o(n) vertices of the graph can be covered with two vertex-disjoint monochromatic cycles of distinct colours, if colourings close to a split colouring are excluded. From this we derive that the whole graph, if large enough, may be covered with 14 vertex-disjoint monochromatic cycles. keywords: monochromatic path partition, monochromatic cycle partition, two-coloured graph MSC: 05C38, 05C55.

1

Introduction

1.1

State of the art

Gerencs´er and Gy´ arf´ as [6] observed the vertex set of any complete graph whose edges are coloured red and blue1 can be partitioned into a red and a blue path. This is fairly easy: just take a maximal set S of vertices that span two paths P1 , P2 , one in each colour, which only meet in one of their endvertices, call this vertex x. One quickly checks that any vertex v ∈ / S can be used to augment S: we can add the edge xv to the path Pi of the same colour, and then go from v on ∗ Supported

by Fondecyt Regular no. 1140766. that a colouring is never meant to be a proper colouring in this paper: any assignment of colours red any blue to the edges will do. 1 Note

1

reversely through P3−i . It is a long-standing conjecture that this phenomenon carries over to arbitrarily many colours. Conjecture 1.1 (Gy´ arf´ as [8]). Let G be a complete graph whose edges are coloured with r colours. Then G can be partitioned into r monochromatic paths. A stronger conjecture, replacing paths by cycles, had been put forward by Erd˝ os, Gy´ arf´ as and Pyber [5], but was recently disproved by Pokrovskiy [18] for r ≥ 3. (Here, and throughout the paper, a cycle is allowed to consist of a single vertex or an edge, or to be totally empty.) In the case r = 2, however, the stronger result with cycles does hold (even with cycles of distinct colours). This used to be known as Lehel’s conjecture, and was shown for all n by Bessy and Thomass´e [3], after having been proved for large values of n by Luczak, R¨ odl and Szemer´edi [16] and by Allen [1]. Theorem 1.2 (Bessy and Thomass´e [3]). Let G be a complete graph whose edges are coloured red and blue. Then G can be partitioned into a red and a blue cycle. Together with Conlon, the second author showed in [4] that Theorem 1.2 literally extends to 2-local colourings: those are colourings with any number of colours, where each vertex is incident with at most two colours. For arbitrary r, the best known bound on the number of vertex-disjoint cycles needed to cover the r-coloured complete graph Kn is 100r log r, if n is large, this bound is due to Gy´ arf´as, Ruszink´o, S´ark¨ozy and Szemer´edi [10]. For r = 3, the same authors show in [12] that there is a partition of all but o(n) vertices of Kn into 3 or less monochromatic cycles. From this they deduce that 17 cycles partition the whole graph. If one aims for similar results in complete bipartite graphs, it is reasonable to assume these are balanced, i.e. the two partition classes have the same size. As observed by several authors, an obstruction for partitions of two-coloured balanced complete bipartite graphs into two paths/cycles is a certain type of colouring, which we will now describe. For any bipartite graph with partition classes U , V , call a red/blue colouring of E(G) a split colouring if there are partitions U = A ∪ B and V = C ∪ D such that all edges in EG (A, C) ∪ EG (B, D) are blue, and all edges in EG (A, D) ∪ EG (B, C) are red. A split colouring is proper if min{|A| − |C|, |B| − |D|} ≥ 2. It is easy to see that a balanced complete bipartite graph with a proper split colouring cannot be partitioned into two monochromatic paths (even if the paths are allowed to have the same colour). That the converse is also true was shown by Pokrovskiy [18], improving an earlier result of Gy´arf´as and Lehel [7, 9] (they allowed one uncovered vertex). Theorem 1.3 (Pokrovskiy [18]). Let G be a balanced complete bipartite graph whose edges are coloured red and blue. If the colouring is not a proper split colouring, then G can be partitioned into a red and a blue path. It is not difficult to check that the vertices of any balanced complete bipartite graph with a proper split colouring can be partitioned into three monochromatic paths, or cycles. Haxell [13] proved that any r-edge coloured balanced complete bipartite graph can be partitioned into O((r log r)2 ) monochromatic cycles, and if r = 3, 2

then 1695 monochromatic cycles suffice. In [14] we improve this number to 18. Finally, we mention that S´ark˝ozy [19] conjectured that any r-edge coloured graph G can be partitioned into rα(G) monochromatic cycles: though this is false by Pokrovskiy’s counterexample mentioned above, it is asymptotically true as shown by Balogh et al. [2].

1.2

Our Contribution

We investigate monochromatic path or cycle partitions into complete multipartite graphs with more than two partition classes. We always assume that none of the partition classes is empty, e.g., a complete tripartite graph is assumed to be non bipartite. We call a multipartite graph fair if no partition class contains more than half of the vertices of the graph. Note that a complete bipartite graph is fair if and only if it is balanced, and, more generally, a complete multipartite graph is fair if and only if it admits a Hamilton cycle. We show an extension of Theorem 1.3 to multipartite graphs with more than two partition classes under the necessary restriction of fairness. It is interesting that for these graphs, there is no analogue for the exceptional case of the split colouring. Theorem 1.4. Let G be a fair complete k-partite graph, with k ≥ 3, whose edges are coloured red and blue. Then G can be partitioned into a red and a blue path. It seems plausible that Theorem 1.4 can be strengthened such that instead of two paths one can partition G into a path and a cycle. Indeed, we shall see this is true if we allow for at most one vertex to be uncovered. Corollary 1.5. Let G be a fair complete k-partite graph, with k ≥ 3, whose edges are coloured red and blue. Then all but at most one vertex of G can be partitioned into a monochromatic path and a monochromatic cycle of distinct colours. We prove Theorem 1.4 and Corollary 1.5 in Section 2. The main tool for this proof is Lemma 2.2, which is shown in Section 3. It is natural to ask whether Theorems 1.3 and 1.4 extend to cycle partitions instead of path partitions. Note that we need to exclude the situation that there is a proper split colouring between a partition class that contains half the vertices of the graph, and the rest of the graph. Clearly, in that case a partition into two monochromatic cycles cannot exist (while a partition into two monochromatic paths is possible, by Theorem 1.4). We show an approximate result for the cycle partition problem in complete multipartite graphs, including the bipartite case. For this, we say that a colouring of the edges of a complete multipartite graph G is δ-close to a split colouring if by deleting at most δ|E(G)| edges we can make G bipartite and the colouring a split colouring. Theorem 1.6. For all δ > 0 there is a an n0 such that the following holds for every fair complete k-partite graph G on n > n0 vertices, with k ≥ 2. If the edges of G are coloured red and blue, and the colouring is not δ-close to a split colouring, then there are two disjoint monochromatic cycles of distinct colours, which together cover all but at most δn vertices of G. 3

It is easy to check that no colouring of a complete multipartite √ graph on n vertices, with at least three partition classes of size greater than 2 δn, can be δ-close to a split colouring. So, for these graphs we can drop the condition on the colouring in Theorem 1.6. Also, notice that any complete multipartite graph on n vertices, with a colouring that is δ-close to a split colouring, contains √ three disjoint monochromatic cycles that together cover all but at most 8 δn vertices of the graph.2 We prove Theorem 1.6 in Section 5. The strategy uses the regularity method, and a well-known technique due to Luczak for blowing up connected matchings of the reduced graph to cycles in the original graph. The existence of the connected matchings in the given circumstances is shown in Lemma 4.2 and Lemma 4.3 of Section 4. Theorem 1.6 is probably not the best possible result. It might be possible to cover all but a constant number of vertices of our multipartite graph. For an open problem in this direction, and more discussion, see Section 7. Using tools of Gy´ arf´ as, [11] and Haxell [13], we derive from Theorem 1.6 that a small finite number of monochromatic cycles is always sufficient to partition a multipartite graph. Theorem 1.7. Let G be a sufficiently large fair complete k-partite graph, whose edges are coloured red and blue. Then G can be partitioned into 14 monochromatic cycles. If k = 2, then G can be partitioned into 12 monochromatic cycles. We prove Theorem 1.7 in Section 6. We believe that probably, the number of cycles can be dropped further, but the point of our result is rather that a reasonable finite number of cycles always suffices. We end the introduction with a useful lemma, which tells us that for all our results for fair k-partite graphs with k ≥ 3, we may restrict our attention to the tripartite case. Lemma 1.8. Every fair k-partite graph G with k ≥ 3 has a spanning induced subgraph which is fair and tripartite. Proof. Assume k ≥ 4 and delete all edges between the smallest two partition classes. If the resulting graph is not fair, then these two classes together contain more than |V (G)|/2 vertices. But then, also the third and the fourth smallest class together have more than |V (G)|/2 vertices, a contradiction. Inductively, the statement follows.

2

Proof of Theorem 1.4

Throughout this section, let G = (V, E) be a fair complete k-partite graph on n vertices whose edges are coloured red and blue. Let V1 , . . . , Vk be the partition classes of G and let ni = |Vi | for i = 1, . . . , k, with n1 ≥ . . . ≥ nk . Our aim is to partition G into a red and a blue path. By Lemma 1.8, we may assume k = 3. 2 In fact, in the graph induced by the edges of the split colouring, we can delete a balanced √ set √ of at most 2 δn vertices so that in the remaining graph H, each vertex has at most δn non-neighbours in the other partition class. Then we split H into three monochromatic balanced bipartite graphs, two in blue, and one in √ red. It is easy to check that each of these three graphs has a cycle covering all but at most 2 δn of its vertices.

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Lemma 2.1. If n1 = n/2, then G can be partitioned into a red and a blue path sharing a common endvertex. Proof. Consider the balanced complete bipartite graph H = (V, EG (V1 , V2 ∪V3 )) whose partition classes are V1 and V2 ∪V3 . By Theorem 1.3, H and thus G can be partitioned into a red and a blue path if the edges of H are not split-coloured. So, we may assume that the edges of H are coloured with a split colouring, that is, there are disjoint non-empty sets A, B ⊆ V1 and C, D ⊆ V2 ∪ V3 with A ∪ B = V1 and C ∪ D = V2 ∪ V3 , such that EG (A, C) ∪ EG (B, D) is entirely coloured blue, and EG (A, D) ∪ EG (B, C) is entirely coloured red. Since V2 , V3 6= ∅ we have that EG (C, D) 6= ∅. Let uv ∈ EG (C, D), say uv is blue. Let Pu (Pv ) be a blue path in H, of maximum even length starting in u (in v). Let P be the blue path Pu uvPv . If |A| ≤ |C|, then |D| ≤ |B|, and thus A ∪ D ⊆ V (P ). Otherwise, B ∪ C ⊆ V (P ). In both cases H[V \ V (P )] is a balanced complete bipartite graph with red edges only. Hence, G can be partitioned into a red and a blue path. It is straightforward that there is an edge joining an endvertex of the red path to an endvertex of the blue path. The following is the main tool for our proof of Theorem 1.4. Lemma 2.2. Assume that n1 ≤ n2 + n3 − 2, or that n1 = n2 + n3 − 1 and n3 > 1. Then G can be partitioned into a red and a blue path sharing a common endvertex. The proof of Lemma 2.2 is the subject of Section 3. Lemma 2.3. Assume that n1 = n2 and n3 = 1. Then G can be partitioned into a red and a blue path sharing a common endvertex. Proof. Say V3 = {z}. Consider the balanced complete bipartite graph H induced by the vertex set V1 ∪ V2 . First, assume that the edges of H are not properly split-coloured. Then by Theorem 1.3, H can be partitioned into a red path R and a blue path B. Clearly, as H is balanced, there are an endvertex r of R and an endvertex b of B which lie in distinct partition classes. We may assume the edge rb to be blue, the other case is analogous. If the edge rz is red, we may extend the red path to include z. Together with the edge bz, this gives the desired partition. Otherwise, if rz is blue, we may extend the blue path to include z. Together with the edge r0 z, where r0 is the second to last vertex on R, this gives the desired partition. So we may assume that H is properly split coloured. That is, there are disjoint non-empty sets A, B ⊆ V1 and C, D ⊆ V2 with A ∪ B = V1 and C ∪ D = V2 , such that EG (A, C) ∪ EG (B, D) is entirely coloured blue, and EG (A, D) ∪ EG (B, C) is entirely coloured red. Now, there are two colour-components, either A ∪ D and B ∪ C, or A ∪ C and B ∪ D which are connected in G via z. We treat the case that there are vertices Let a ∈ A, b ∈ B such that the edges az, bz are red, all other cases can be treated analogously. Choose a longest balanced red path X starting in a, and a longest balanced red path Y starting in b (where balanced means the path should have an even number of vertices). Take a longest blue path Z covering G − (V (X) ∪ V (Y ) ∪ {z}). The latter choice is possible since A ∪ C ⊆ V (X) ∪ V (Y ) or B ∪ D ⊆ V (X) ∪ V (Y ). Thus, G can be partitioned into the red path XzY and the blue path Z. 5

We are now ready for the proof of our first main theorem. Proof of Theorem 1.4. By Lemma 1.8, it suffices to prove our result for k = 3. If either n1 ≤ n2 +n3 −2, or n1 = n2 +n3 −1 and n3 > 1, we may apply Lemma 2.2 and are done, so assume otherwise. Then either n1 = n2 and n3 = 1, in which case we may apply Lemma 2.3 and are done, or n1 ≥ n2 + n3 , which we assume from now on. Since G is fair, we actually have n1 = n2 + n3 , and are thus in conditions to apply Lemma 2.1 to obtain the desired partition. We also make use of the following simple lemma we shall need for the proof of Lemma 2.2 in Section 3. Since the proof is straightforward, we omit it. Lemma 2.4. Assume that ni ≤ n3−i + n3 − 1 for some i ∈ {1, 2}, and let P be any Hamilton path in G. Then either P contains an edge uv ∈ EG (V3−i , V3 ), or ni = n3−i + n3 − 1 and both endvertices of P are in V3−i ∪ V3 . We end this section with the proof of Corollary 1.5. Proof of Corollary 1.5. By Lemma 1.8, we may assume G is tripartite, with partition classes V1 , V2 , V3 . We observe that the lemmas we use to prove Theorem 1.4, Lemmas 2.1, 2.2, and 2.3, yield a partition of G into a red and a blue path, say R and B, that share their last vertex x. Say among all such partitions, R is chosen of maximum length. W.l.o.g. assume x ∈ V1 . If any of the two paths R, B is trivial, or has only one edge, we are done. Note that we may assume all edges between the first two and the last two vertices of R to be blue, and all edges between the first two and the last two vertices of B to be red, as otherwise we are done. In particular, by maximality of R, this implies that the first vertex v1 on B lies in V1 . First assume the first vertex w1 on R does not lie in V1 . Then w1 x is blue, and by maximality of R, we know that w1 v1 is blue, too. Thus we are done. So assume w1 ∈ V1 . By maximality of R, we know w1 sends a blue edge to the second last vertex on B. Let v2 be the second vertex on B. Now if v2 w1 is red we find a red cycle and a blue path covering all but one vertex, and if v2 w1 is blue we find a blue cycle and a red path covering all but one vertex.

3

Proof of Lemma 2.2

This section is devoted to the proof of Lemma 2.2. For notational reasons it will be very convenient to now refrain from the assumption that n1 ≥ n2 . We still keep the convention that V3 is the smallest of the three classes, that is, n3 ≤ min{n1 , n2 }. We thus have to prove the following statement. Assume that ni ≥ n3 for i ∈ {1, 2}, and that either ni ≤ n3−i + n3 − 2, or ni = n3−i + n3 − 1 and n3 > 1. Then G can be partitioned into a red and a blue path sharing a common endvertex. The assumptions of the lemma imply that min{n1 , n2 } ≥ 2. Let v1 ∈ V1 and v2 ∈ V2 be arbitrary, let H = G − {v1 , v2 }, and let n01 , n02 , n03 be the sizes of the partition classes of H. If n01 , n02 , n03 satisfy the statement above, we may apply induction to see that H can be partitioned into a red path R and a blue path B that share a common endvertex. 6

If n01 , n02 , n03 violate the statement above, it must be that min{n01 , n02 } < n03 . As n01 = n1 − 1 and n02 = n2 − 1, we may w.l.o.g. assume that n2 = n3 . So, n1 ≥ n2 = n3 . Let us first discuss the case n1 = n2 = n3 . It must be that n1 = 2, for otherwise n03 ≤ n01 + n02 − 1, n01 ≤ n02 + n03 − 1 and n02 > 1, a contradiction to our assumption that n01 , n02 , n03 violate the statement. Thus n03 = n01 + n02 and so we may apply Lemma 2.1 to obtain a red path R and a blue path B sharing a common endvertex that partition H. Now assume n1 > n2 = n3 . Hence, n01 ≥ n03 > n02 ≥ 1. Again we have 0 n1 ≤ n02 + n03 − 1, n03 ≤ n01 + n02 − 1. Since n01 , n02 , n03 violate the statement, it must be that n02 = 1. As n2 ≥ n3 and n02 = n2 − 1, we have n2 = 2 and n03 = n3 = 2. Therefore n1 = 3 and thus n01 = 2. So, we can apply Lemma 2.3 to H and obtain a red path R and a blue path B sharing a common endvertex that partition H. Summing up, we may inductively assume Lemma 2.2 to hold for H. Hence, H can be partitioned into a red path R and a blue path B, both possibly trivial, that share a common endvertex. Let R = (r1 , . . . , rs , x) and B = (b1 , . . . , bt , x) where R and B have only x in common. Throughout the proof we suppose for contradiction that G cannot be partitioned into a red and a blue path that share a common endvertex. In several cases treated below we make use of the following simple fact. Assume we can partition G into a red path R0 and a blue path B 0 such that one of these paths has its endvertices in distinct partition classes. Then one of R0 , B 0 can be extended such that the paths have exactly one vertex in common, namely an endvertex of both paths. The same holds if an endvertex of one path is in a distinct partition class than an endvertex of the other path. For the remainder of the proof, we assume w.l.o.g. that v1 v2 is red. At this point, we advise the reader to get his coloured pencils ready. Claim 3.1. Neither R nor B is trivial. (That is R 6= (x) 6= B.) Proof. We first suppose that the path R is trivial, that is, B covers H. W.l.o.g. assume b1 ∈ / V1 . If the edge b1 v1 is red, then G can be partitioned into the red path (v2 , v1 , b1 ) and the blue path (b1 , . . . , bt , x), a contradiction. Otherwise, G can be partitioned into the red path (v2 , v1 ) and the blue path (v1 , b1 , . . . , bt , x), another contradiction. Now suppose that B is trivial, that is, R covers H. Again, we may w.l.o.g. assume that r1 ∈ / V1 . The edge r1 v1 is blue, since otherwise the red path (v2 , v1 , r1 , . . . , rs , x) covers G, a contradiction. If r1 ∈ V3 , then r1 v2 ∈ E. In this case, however, G is covered by the red path (v1 , v2 , r1 , . . . , rs , x), if r1 v2 is red. If r1 v2 is blue, G can be partitioned into the red path (r2 , . . . , rs , x) and the blue path (v1 , r1 , v2 ). As both cases are contradictory, r1 ∈ / V3 and so r1 ∈ V2 . By symmetry, x ∈ / V3 . First we suppose that x ∈ V1 . Then xv2 , xr1 ∈ E. Like above, xv2 must be blue. But then r1 x is red, for otherwise G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , r1 , x, v2 ), a contradiction (because of the observation stated before the claim we are presently proving). We may pick i ∈ {2, . . . , s} such that ri ∈ V3 . If the edge v1 ri is red, G can be partitioned into the red path (v2 , v1 , ri , . . . , rs , x, r1 , . . . , ri−1 ), a contradiction. Thus, by symmetry, both v1 ri and v2 ri are blue. This means G can be partitioned into the red path (ri+1 , . . . , rs , x, r1 , . . . , ri−1 ) and the blue path (v1 , ri , v2 ), a contradiction. 7

So, x ∈ V2 . Then v1 x ∈ E, and this edge must be blue, as it is interchangeable with v1 r1 . Consider the edge v2 r2 : it must be blue, else G can be partitioned into the red path (v1 , v2 , r2 , . . . , rs , x) and the blue path (v1 , r1 ). Thus, the edge r2 x is red, since otherwise G can be partitioned into the red path (r3 , . . . , rs−1 , rs ) and the blue path (v2 , r2 , x, v1 , r1 ). But this is a contradiction: if the red path (r3 , . . . , rs ) is non-empty, rs ∈ / V2 and so v2 rs ∈ E. By Lemma 2.4, we may pick i ∈ {2, . . . , s} such that ri ∈ V3 and ri−1 ∈ V1 or ri+1 ∈ V1 . Note that we may apply Lemma 2.4 even if n0i = n03−i + n03 − 1, for some i ∈ {1, 2}, since r1 ∈ V2 . Let us assume that ri−1 ∈ V1 , the other case is similar. If ri v1 is red, then G can be partitioned into the red path (v2 , v1 , ri , . . . , rs , x, r2 , . . . , ri−1 ) and the blue path (r1 ), a contradiction. Similarly, ri v2 cannot be red. So, both ri v1 and ri v1 are blue. But now G can be partitioned into the red path (ri+1 , . . . , rs , x, r2 , . . . , ri−1 ) and the blue path (v2 , ri , v1 , r1 ), which is contradictory. This completes the proof of Claim 3.1. Over the next few claims, we deal with the case that x ∈ V3 . Claim 3.2. If x ∈ V3 , then r1 ∈ / V3 . Proof. Suppose x, r1 ∈ V3 . Consider the edges xv1 and xv2 . If any of these edges is red, then the respective path (r1 , . . . , rs , x, v1 , v2 ) or (r1 , . . . , rs , x, v2 , v1 ) is red, and together with the blue path (b1 , . . . , bt ) covers G. Thus xv1 and xv2 are blue. If either of r1 v1 , r1 v2 is red, we may simply extend R from r1 to v1 and v2 . So these two edges are blue, too. Observe that one of the blue paths (b1 , . . . , bt , x, v1 , r1 , v2 ), (b1 , . . . , bt , x, v2 , r1 , v1 ) has its endpoints in different partition classes. Together with the red path (r2 , . . . , rs ), this blue path covers G, a contradiction. Claim 3.3. Let i ∈ {1, 2}. If x ∈ V3 , then not both r1 and rs lie in Vi . Proof. Because of symmetry, it is enough to prove this claim for i = 1. So for contradiction suppose that x ∈ V3 and r1 , rs ∈ V1 . As above, we see that xv1 and r1 v2 are blue. Also, rs v2 is blue, since otherwise G can be partitioned into the red path (r1 , . . . , rs , v2 , v1 ) and the blue path (v1 , x, bt , . . . , b1 ). We claim that b1 ∈ V1 . (1) Suppose b1 ∈ / V1 . Then xv2 must be blue, since otherwise G can be partitioned into the red path (r1 , . . . , rs , x, v2 , v1 ) and the blue path (b1 , . . . , bt ). As v1 b1 ∈ E, this edge must be red, for otherwise G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v2 , x, bt , . . . , b1 , v1 ). Thus, r1 b1 is blue, else G can be partitioned into the red path (rs , . . . , r1 , b1 , v1 , v2 ) and the blue path (v2 , x, bt , . . . , b2 ). Hence, G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , x, bt , . . . , b1 , r1 , v2 ), a contradiction. This proves (1). Hence, the edge b1 v2 exists, and b1 v2 is red,

(2)

since otherwise G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v1 , x, bt , . . . , b1 , v2 ). If t = 1, xb1 is blue by definition. If t ≥ 2 and xb1

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is red, G can be partitioned into the red path (r1 , . . . , rs , x, b1 , v2 , v1 ) and the blue path (b2 , . . . , bt ) (note that by (1) we know that b2 ∈ / V1 ). Thus, xb1 is blue.

(3)

bt ∈ V1 .

(4)

We now show that Indeed, suppose otherwise. Then the edge bt r1 exists. If bt r1 is blue, then G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , x, b1 , . . . , bt , r1 , v2 ). This is contradictory since v1 ∈ V1 and v2 ∈ V2 . So bt r1 is red. Since xbt ∈ E, we have bt ∈ V2 . Thus, v1 bt ∈ E, and this edge must be blue: otherwise G can be partitioned into the red path (x, rs , . . . , r1 , bt , v1 , v2 ) and the by (3) blue path (x, b1 , . . . , bt ). Moreover, xv2 must be blue, else G can be partitioned into the red path (r1 , . . . , rs , x, v2 , v1 ) and the blue path (v1 , bt , . . . , b1 , ). But this means G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v2 , x, v1 , bt , . . . , b1 ), a contradiction. This proves (4). By (4), the edge bt v2 exists. However, this edge cannot be blue, as then G can be partitioned into the red path (rs , . . . , r1 ) and the blue path (rs , v2 , bt , . . . , b1 , x, v1 ). So, bt v2 is red. We next show that xr1 is red. (5) For contradiction, suppose that xr1 is blue. Then, in particular, s ≥ 2, since xrs is red. The edge r2 v1 must be blue, as otherwise G can be partitioned into the red path (v2 , v1 , r2 , . . . , rs ) and the blue path (v2 , r1 , x, bt , . . . , b1 ). Moreover, bt r2 is blue, as follows. Suppose bt r2 is red. If t = 1, G can be partitioned into the red path (v1 , v2 , bt , r2 , . . . , rs , x) and the blue path (r1 , x). Otherwise, bt−1 exists and does not belong to V1 . Thus, G can be partitioned into the red path (v1 , v2 , bt , r2 , . . . , rs ) and the blue path (r1 , x, b1 , . . . , bt−1 ). As both is contradictory, we see that bt r2 is blue. But now the red path (r3 , . . . , rs ) and the blue path (rs , v2 , r1 , x, v1 , r2 , bt , . . . , b1 ) partition G, a contradiction. This proves (5). We now apply Lemma 2.4 to H, to see that there is an edge uv in R or B with u ∈ V3 and v ∈ V2 . Note that we may apply Lemma 2.4 even if n0i = n03−i + n03 − 1, for some i ∈ {1, 2}, since r1 ∈ V1 by assumption. Since rs , bt ∈ V1 , we know that v ∈ / {rs , bt }, and thus u 6= x. We claim that u, v ∈ V (B). (6) Indeed, suppose u, v ∈ V (R). We assume v lies between u and x on R, say u = ri and v = ri+1 , the other case is similar. It must be that ri v2 is blue, since otherwise G can be partitioned into the red path (v1 , v2 , ri , ri−1 , . . . , r1 , x, rs , . . . , ri+1 ) and the blue path (b1 , . . . , bt ). Similarly, the edges ri v1 and ri+1 v1 are blue. Since ri+1 ∈ V2 and b1 ∈ V1 , ri+1 b1 ∈ E. If ri+1 b1 is red, then G can be partitioned into the red path (v1 , v2 , b1 , ri+1 , . . . , rs , x, r1 , . . . , ri ) and the blue path (b2 , . . . , bt ) (recall that b1 v2 is red by (2)). However, if ri+1 b1 is blue, then G can be partitioned into the red path (ri+2 , . . . , rs , x, r1 , . . . , ri−1 ) and the blue path (v2 , ri , v1 , ri+1 , b1 , . . . , bt ). This proves (6).

9

Say v lies between u and x on B, say u = bi and v = bi−1 , the other case is similar. We now show that r1 bi is blue. (7) Indeed, suppose that the edge r1 bi is red. Thus bi v1 must be blue: otherwise, G can be partitioned into the red path (v2 , v1 , bi , r1 , . . . , rs ) and the blue path (bi−1 , . . . , b1 , x, bt , . . . , bi+1 ). Similarly, bi v2 must be blue. Hence bi−1 v1 must be red, as otherwise G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v2 , bi , v1 , bi−1 , . . . , b1 , x, bt , . . . , bi+1 ). Supposing bi−1 r1 is red, G can be partitioned into the red path (v2 , v1 , bi−1 , r1 , . . . , rs ) and the blue path (bi , . . . , bt , x, b1 , . . . , bi−2 ). Thus, the edge bi−1 r1 is blue, and so G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , bi , v2 , r1 , bi−1 , . . . , b1 , x, bt , . . . , bi+1 ). This proves (7). Now we see that v1 bi−1 must be red: otherwise, G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v2 , r1 , bi , . . . , bt , x, b1 , . . . , bi−1 , v1 ). Also, the edge r1 bi−1 is blue, as otherwise we can cover G with the red path (v2 , v1 , bi−1 , r1 , . . . , rs ) and the blue path (bi , . . . , bt , x, b1 , . . . , bi−2 ). Therefore, bi v1 is red, as otherwise G can be partitioned into the red path (r2 , . . . , rs ) together with the blue path (v2 , r1 , bi−1 , . . . , b1 , x, bt , . . . , bi , v1 ). This implies that rs bi−1 is red, else G could be covered by the red path (r2 , . . . , rs−1 ) and the blue path (v1 , x, bt , . . . , bi , r1 , v2 , rs , bi−1 , . . . , b1 ). But then we can cover G with the red path (v1 , bi−1 , rs , . . . , r2 ) and the blue path (bi−2 , . . . , b1 , x, v2 , r1 , bi , . . . , bt ) (here we use (3) and (7)), giving the final contradiction. Claim 3.4. Let i ∈ {1, 2}. If x ∈ V3 , then not both r1 ∈ Vi and rs ∈ V3−i hold. Proof. For symmetry, we only need to treat the case i = 1. So suppose r1 ∈ V1 and rs ∈ V2 . Then rs v1 is blue, (8) r2 v1 is blue, since otherwise G can be partitioned into the red path (r1 , . . . , rs , v1 , v2 ) and the blue path (b1 , . . . , bt , x). Thus, r1 rs is red,

(9)

since otherwise G can be partitioned into the red path (r2 , . . . , rs−1 ) and the blue path (v2 , r1 , rs , v1 , x, bt , . . . , b1 ). Let us show that b1 ∈ V1 . (10) Suppose otherwise. Then r1 b1 ∈ E, and this edge must be red: if it was blue, G could be covered by the red path (r2 , . . . , rs ) together with the blue path (rs , v1 , x, bt , . . . , b1 , r1 , v2 ). Moreover, v1 b1 ∈ E, and this edge must be blue, since otherwise G can be partitioned into the red path (x, rs , . . . , r1 , b1 , v1 , v2 ) and the blue path (x, bt , . . . , b2 ). Also, xv2 is blue, else G can be partitioned into the red path (r1 , . . . , rs , x, v2 , v1 ) and the blue path (b1 , . . . , bt ). Now, however, G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v2 , x, bt , . . . , b1 , v1 ), a contradiction. This proves (10). By (10), we know that v2 b1 ∈ E, and this edge must be red, as otherwise G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (rs , v1 , x, bt , . . . , b1 , v2 ) (recall that rs v1 is blue by (8)). Thus, rs b1 is blue, else

10

G can be partitioned into the red path (r1 , . . . , rs , b1 , v2 , v1 ) and the blue path (v1 , x, bt , . . . , b1 ). Let j ∈ {1, 2} such that nj ≥ n2−j . Hence nj ≤ n2−j + n3 − 2, so n3 ≥ 2. Thus, there is a vertex y ∈ V3 \ {x}. We show that y ∈ V (B).

(11)

Suppose otherwise, say y = ri . Then, ri v2 is blue, as else G can be partitioned into the red path (v1 , v2 , ri , . . . , r2 , r1 , rs , . . . , ri+1 ) (recall that r1 rs is red by (9)) and the blue path (b1 , . . . , bt , x). Similarly, ri v1 is blue. But now G can be partitioned into the red path (ri−1 , . . . , r1 , rs , . . . , ri+1 ) and the blue path (v2 , ri , v1 , x, bt , . . . , b1 ). This proves (11). So say y = bi . Thus, v2 bi is red, otherwise G can be partitioned into the red path (r1 , . . . , rs−1 ) and the blue path (v2 , bi , . . . , bt , x, v1 , rs , b1 , . . . , bi−1 ). The edge r1 bi is red, too, else G can be partitioned into the red path (r2 , . . . , rs−1 ) and the blue path (v2 , r1 , bi , . . . , bt , x, v1 , rs , b1 , . . . , bi−1 ). (For this, note that rs−1 ∈ / V2 .) But now G can be partitioned into the red path (v2 , bi , r1 , . . . , rs−1 ) and the blue path (bi+1 , . . . , bt , x, v1 , rs , b1 , . . . , bi−1 ), a contradiction. This finishes the proof of the claim. Putting Claims 3.2, 3.3 and 3.4 together, and noting that if x ∈ V3 , then rs ∈ / V3 , we obtain the following assertion. Claim 3.5. x ∈ / V3 . We now turn to the case that x ∈ / V3 . We first show an auxiliary claim. Claim 3.6. Let i ∈ {1, 2}. If x ∈ Vi then the edges xv3−i , rs vi are blue, and r1 ∈ / V3 . Furthermore, if r1 ∈ Vj for some j ∈ {1, 2}, then r1 v3−j is blue. Proof. For symmetry, it is enough to show this claim for i = 1. So, assume x ∈ V1 . Then, the edge xv2 is present in G. If xv2 is red, then the red path (v1 , v2 , x, rs , . . . , r1 ) and the blue path (b1 , . . . , bt ) together cover G. So we know that xv2 is blue. The edge rs v1 is present in G as rs ∈ / V1 . If this edge is red, G can be partitioned into the red path (r1 , . . . , rs , v1 , v2 ) and the blue path (x, b1 , . . . , bt ), a contradiction. Thus, rs v1 is blue. Similarly, neither of the edges r1 v1 and r1 v2 , if present, can be red. On the other hand, not both r1 v1 and r1 v2 can be present and blue, else G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , r1 , v2 , x, bt , . . . , b1 ). Thus, either r1 v1 or r1 v2 is absent from G, which implies that r1 ∈ / V3 . This proves the claim. Claim 3.7. Let i ∈ {1, 2}. If x ∈ Vi then r1 ∈ / V3−i . Proof. For symmetry, it is enough to show this claim for i = 1. So, for contradiction, assume x ∈ V1 and r1 ∈ V2 . By Claim 3.6, the edges xv2 , rs v1 and r1 v1 are blue. We show first that b1 ∈ / V1 . (12) Suppose otherwise. Then clearly r1 b1 is red, since otherwise G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , r1 , b1 , . . . , bt , x, v2 ). 11

Thus the edge v2 b1 is blue, otherwise G can be partitioned into the red path (v1 , v2 , b1 , r1 , . . . , rs ) and the blue path (x, bt , . . . , b2 ). A symmetric argument shows that r1 x is red. Moreover, rs b1 is red, for otherwise G can be partitioned into the red path (r2 , . . . , rs−1 ) and the blue path (r1 , v1 , rs , b1 , . . . , bt , x, v2 ). Also, the edge bt v1 is red, as otherwise G can be partitioned into the red path (r2 , . . . , rs−1 ) and the blue path (r1 , v1 , bt , . . . , b1 , v2 , x). Hence, b1 bt is blue, else G can be partitioned into the red path (v2 , v1 , bt , b1 , r1 , . . . , rs , x) and the blue path (b2 , . . . , bt−1 ). Let y ∈ V3 \ {x}. First suppose that y = ri for some 1 ≤ i ≤ s. Thus both v1 ri and v2 ri are blue: if v1 ri is red, say, G can be partitioned into the red path (v2 , v1 , ri , . . . , r1 , x, rs , . . . , ri−1 ) and the blue path (b1 , . . . , bt ), a contradiction. (Note that b1 ∈ V1 and bt ∈ / V1 .) But this means G can be partitioned into the red path (ri+1 , . . . , rs , x, r1 , . . . , ri−1 ) and the blue path (v1 , ri , v2 , b1 , . . . , bt ). Thus y = bi for some 1 ≤ i ≤ t. Hence, v1 bi is red, as otherwise G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (r1 , v1 , bi , . . . , b1 , v2 , x, bt , . . . , bi+1 ). So, the edge r1 bi must be blue, for otherwise G can be partitioned into the red path (v1 , bi , r1 , . . . , rs ) and the blue path (bi+1 , . . . , bt , x, v2 , b1 , . . . , bi−1 ). Consequently, G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , r1 , bi , bi+1 , . . . , bt , x, v2 , b1 , . . . , bi−1 ). This finishes the proof of (12). Next, we show that b1 ∈ / V3 .

(13)

Suppose otherwise. Then r1 b1 ∈ E and this edge must be red, since otherwise G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , r1 , b1 , . . . , bt , x, v2 ). In the case that v1 b1 is red, G can be partitioned into the red path (v2 , v1 , b1 , r1 , . . . , rs ) and the blue path (x, bt , . . . , b1 ). So v1 b1 is blue, thus G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v2 , x, bt , . . . , b1 , v1 ). This finishes the proof of (13). Putting (12) and (13) together, we see that b1 ∈ V2 . Then b1 v1 must be red, as otherwise G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v2 , x, bt , . . . , b1 , v1 ). Hence xb1 is blue, since otherwise G can be partitioned into the red path (r1 , . . . , rs , x, b1 , v1 , v2 ) and the blue path (b2 , . . . , bt ). Moreover, bt v1 is red, otherwise G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v2 , x, b1 , . . . , bt , v1 ). We claim that rs , bt ∈ V2 . (14) For this, first assume the edge rs bt does exist. If rs bt is red, G can be partitioned into the red path (r1 , . . . , rs , bt , v1 , v2 ) and the blue path (x, b1 , . . . , bt ). If rs bt is blue, however, G can be partitioned into the red path (r1 , . . . , rs−1 ) and the blue path (v1 , rs , bt , . . . , b1 , x, v2 ). This shows that rs bt does not exist. Since x ∈ V1 , we get that rs , bt ∈ Vi for some i ∈ {2, 3}. Now, if rs , bt ∈ V3 , then rs v2 ∈ E. If rs v2 is red, G can be partitioned into the red path (r1 , . . . , rs , v2 , v1 ) and the blue path (x, bt , . . . , b1 ). But if rs v2 is blue, G can be partitioned into the red path (r1 , . . . , rs−1 ) and the blue path (v1 , rs , v2 , x, bt , . . . , b1 ). This proves (14). 12

Since r1 ∈ V2 , we have that r2 bt ∈ E. We show that r2 bt is blue.

(15)

Suppose otherwise. Then also r1 bt−1 is red, as else G can be partitioned into the red path (v2 , v1 , bt , r2 , . . . , rs , x) and the blue path (b1 , . . . , bt−1 , r1 ). Hence, the edge v2 bt−1 is blue, for otherwise G can be partitioned into the red path (bt , v1 , v2 , bt−1 , r1 , . . . , rs , x) and the blue path (b1 , . . . , bt−2 ). So, rs bt−1 is blue, else G can be partitioned into the red path (v2 , v1 , bt , r2 , . . . , rs , bt−1 , r1 ) and the blue path (x, b1 , . . . , bt−2 ). Now, however, G can be partitioned into the red path (bt , r2 , . . . , rs−1 ) and the blue path (r1 , v1 , rs , bt−1 , v2 , x, b1 , . . . , bt−2 ). This proves (15). Observe that r2 rs is red, (16) for otherwise G can be partitioned into the red path (rs−1 , . . . , r3 ) and the blue path (r1 , v1 , rs , r2 , bt , . . . , b1 , x, v2 ). We now apply Lemma 2.4 to H, to see that there is an edge uv on R or on B with u ∈ V3 and v ∈ V1 . We may apply Lemma 2.4 even if n0i = n03−i + n03 − 1, for some i ∈ {1, 2}, since r1 ∈ V2 . We claim that u ∈ V (B). (17) Suppose otherwise, i.e. assume u = ri for some 2 ≤ i ≤ t. We discuss the case v = ri+1 only, the other case is similar. Observe that not both v1 ri and v2 ri can be blue, for otherwise the red path (ri−1 , . . . , r2 , rs , . . . , ri+1 ) and the blue path (r1 , v1 , ri , v2 , x, bt , . . . , b1 ) cover G. Say v1 ri is red, the other case can be resolved similarly. If r1 x is blue, then we can cover G with the red path (v2 , v1 , ri , . . . , r1 , rs , rs−1 , . . . , ri+1 ) and the blue path (r1 , x, bt , . . . , b1 ). Hence, r1 x is red, and so G can be partitioned into the red path (v2 v1 , ri , ri−1 , . . . , r1 , x, rs , . . . , ri+1 ) and the blue path (b1 , . . . , bt ). This proves (17). Say u = bi for some 2 ≤ i ≤ t. We assume v = bi−1 , as the other case is similar. If v1 bi is blue, v2 r2 must be red: otherwise, G can be partitioned into the red path (r3 , . . . , rs ) and the blue path (r1 , v1 , bi , . . . , bt , r2 , v2 , x, b1 , . . . , bi−1 ). But then G can be partitioned into the red path (v2 , r2 , . . . , rs ) and the blue path (r1 , v1 , bi , . . . , bt , r2 , v2 , x, b1 , . . . , bi−1 ). Thus, v1 bi must be red. Hence, r1 bi is blue, for otherwise we can cover G with the red path (v2 , v1 , bi , r1 , . . . , rs ) and the blue path (bi+1 , . . . , bt , b1 , . . . , bi−1 ). So, the edge r2 v2 is blue, as otherwise G can be partitioned into the red path (v1 , v2 , r2 , . . . , rs ) and the blue path (r1 , bi , . . . , bt , x, b1 , . . . , bi−1 ). But then G can be partitioned into the red path (r3 , . . . , rs−1 ) and the blue path (rs , v1 , r1 , bi , . . . , bt , r2 , v2 , x, b1 , . . . , bi−1 ), yielding the final contradiction. Claim 3.8. Let i ∈ {1, 2}. If x ∈ Vi then r1 ∈ / Vi . Proof. Because of symmetry, we only show the claim for i = 1. So suppose x ∈ V1 and r1 ∈ V1 . By Claim 3.6, the edges rs v1 , v2 x and r1 v2 are blue. We first show that b1 ∈ / V1 . (18) Suppose otherwise. Then the edge rs b1 is red, for otherwise G can be partitioned into the red path (r2 , . . . , rs−1 ) and the blue path (v1 , rs , b1 , . . . , bt , x, v2 ,

13

r1 ). Thus, the edge b1 v2 is blue, else G can be partitioned into the red path (v1 , v2 , b1 , rs , . . . , . . . , r1 ) and the blue path (b2 , . . . , bt , x). Moreover, r2 v1 is blue, since otherwise G can be partitioned into the red path (v1 , r2 , . . . , rs ) and the blue path (r1 , v2 , b1 , . . . , bt , x). Hence, r2 b1 must be red, as otherwise G can be partitioned into the red path (rs−1 , . . . , r3 ) and the blue path (rs , v1 , r2 , b1 , . . . , bt , x, v2 , r1 ). Thus, v1 b2 must be red, as otherwise G can be partitioned into the red path (rs−1 , . . . , r2 , b1 ) and the blue path (rs , v1 , b2 , . . . , bt , x, v2 , r1 ). Applying Lemma 2.4 to H, we see that there is an edge uv in R or B with u ∈ V3 and v ∈ V2 . Note that we may apply Lemma 2.4 even if n0i = n03−i +n03 −1, for some i ∈ {1, 2}, since r1 ∈ V1 by assumption. First assume u ∈ V (R), i.e. u = ri for some 2 ≤ i ≤ t. Say v = ri−1 , the other case is similar. If ri v1 is red, then G can be partitioned into the red path (v1 , ri , . . . , r2 , b1 , rs , . . . , ri+1 ) and the blue path (r1 , v2 , x, bt , . . . , b2 ). Thus ri v1 is blue and, similarly, ri−1 v1 is blue. Either ri or ri−1 is adjacent to b2 , say ri b2 ∈ E (the other case is analogous). If ri b2 is red, G can be partitioned into the red path (v1 , b2 , ri , . . . , rs , b1 , r2 , . . . , ri−1 ) and the blue path (r1 , v2 , x, bt , . . . , b3 ). Otherwise, G can be partitioned into the red path (ri−1 , . . . , r2 , b1 , rs , . . . , ri+1 ) and the blue path (r1 , v2 , x, bt , . . . , b2 , ri , v1 ). However, both is contradictory, and we may thus assume that u ∈ V (B). So, u = bi for some 2 ≤ i ≤ t. We discuss the case v = bi−1 only, as the case v = bi+1 can be treated similarly. Now, the edge bi v1 is red, since otherwise G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v1 , bi , bi+1 , . . . , bt , x, v2 , b1 , . . . , bi−1 ). Similarly, bi−1 v1 is red. Clearly r1 bi ∈ E, and this edge must be blue. Otherwise, G can be partitioned into the red path (v1 , bi , r1 , . . . , rs ) and the blue path (bi−1 , . . . , b1 , v2 , x, bt , . . . , bi+1 ). Similarly, r1 bi−1 is blue. It is clear that r2 bi ∈ E or r2 bi−1 ∈ E, say r2 bi ∈ E. If the edge r2 bi is blue, then we can cover G with the red path (r3 , . . . , rs ) and the blue path (v1 , r2 , bi , r1 , bi−1 , . . . , b1 , v2 , x, bt , . . . , bi+1 ). Otherwise, we can partition G into the red path (v1 , bi , r2 , . . . , rs ) and the blue path (r1 , bi−1 , . . . , b1 , v2 , x, bt , . . . , bi+1 ). This finishes the proof of (18). Next, we show that b1 ∈ / V3 .

(19)

Suppose otherwise. Then v1 b1 ∈ E and this edge must be red, otherwise G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v1 , b1 , . . . , bt , x, v2 ). Now assume rs ∈ V2 . In that case, if rs b1 is red, we cover G with the red path (v2 , v1 , b1 , rs , . . . , r1 ) and the blue path (b2 , . . . , bt , x). If rs b1 is blue, we cover G with the red path (rs−1 , . . . , r1 ) and the blue path (v1 , rs , b1 , b2 , . . . , bt , x, v2 ). Thus rs ∈ / V2 and hence rs ∈ V3 . But now, the edge rs v2 cannot be red, because of the red path (r1 , . . . , rs , v2 , v1 ) and the blue path (b1 . . . , bt , x). It also cannot be blue, because of the red path (r1 , . . . , rs−1 ) and the blue path (b1 . . . , bt , x, v2 , rs , v1 ). This finishes the proof of (19). By (18) and (19), we know that b1 ∈ V2 . Thus b1 v1 ∈ E and this edge must be red, as otherwise G can be partitioned into the red path (r1 , . . . , rs ) and the blue path (v1 , b1 , . . . , bt , x, v2 ). So, b1 r1 is blue, for otherwise G can be partitioned into the red path (v2 , v1 , b1 , r1 , . . . , rs ) and the blue path (x, bt , . . . , b1 ). 14

Thus, bt v1 is red, since otherwise G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , bt , x, v2 , r1 , b1 , . . . , bt−1 ). This implies that the edge bt r1 is blue, as otherwise G can be partitioned into the red path (v2 , v1 , bt , r1 , . . . , rs , x) and the blue path (bt−1 , . . . , b1 ). Moreover, r1 rs is red, for otherwise G can be partitioned into the red path (r2 , . . . , rs−1 ) and the blue path (v1 , rs , r1 , v2 , x, bt , . . . , b1 ). Also, xb1 is blue, else G can be partitioned into the red path (v2 , v1 , b1 , x, rs , . . . , r1 ) and the blue path (bt−1 , . . . , b2 ). Applying Lemma 2.4 to H, we see that there is an edge uv in R or B with u ∈ V3 and v ∈ V1 . Note that we may apply Lemma 2.4 even if n0i = n03−i +n03 −1, for some i ∈ {1, 2}, since we know that b1 ∈ V2 . First assume u ∈ V (R), i.e. u = ri for some 2 ≤ i ≤ t. We may suppose that v = ri−1 , the other case is similar. Then ri v1 is blue, for otherwise G can be partitioned into the red path (v2 , v1 , ri , . . . , rs , r1 , . . . , ri−1 ) and the blue path (x, bt , . . . , b1 ). Similarly, ri v2 is blue. But then G can be partitioned into the red path (ri−1 , . . . , r1 , rs , . . . , ri+1 ) and the blue path (v1 , ri , v2 , x, bt , . . . , b1 ), a contradiction. Therefore, u ∈ V (B). Say u = bi for some 2 ≤ i ≤ t. Say v = bi−1 , the other case is similar. Thus, bi v1 is red, for otherwise G can be partitioned into the red path (r2 , . . . , rs ) and the blue path (v1 , bi , . . . , b1 , r1 , v2 , x, bt , . . . , bi+1 ). Hence, the edge r1 bi is blue, for otherwise G can be partitioned into the red path (v2 , v1 , bi , r1 , . . . , rs ) and the blue path (bi−1 , . . . , b1 , x, bt , . . . , bi+1 ). If rs ∈ V3 , then rs v2 ∈ E. But then rs v2 must be blue, since otherwise G can be partitioned into the red path (r1 , . . . , rs , v2 , v1 ) and the blue path (x, bt , . . . , b1 ). This implies that G can be partitioned into the red path (r1 , . . . , rs−1 ) and the blue path (v1 , rs , v2 , x, bt , . . . , b1 ). So, rs ∈ / V3 , and hence rs ∈ V2 . Thus, the edge rs bi exists and it must be red, for otherwise G can be partitioned into the red path (r2 , . . . , rs−1 ) and the blue path (v1 , rs , bi , . . . , bt , x, v2 , r1 , b1 , . . . , bi−1 ), where v1 ∈ V1 and r2 ∈ / V1 . Hence, G can be partitioned into the red path (v2 , v1 , bi , rs , . . . , r1 ) and the blue path (bi+1 , . . . , bt , x, b1 , . . . , bi−1 ), a contradiction. Putting Claims 3.5, 3.7 and 3.8 together, we arrive at the final contradiction and thus complete the proof of Lemma 2.2.

4

Connected matchings

Let G be a graph whose edges are coloured red or blue. A monochromatic matching of G is called connected if it is contained in a connected component of the subgraph induced by the edges of the corresponding colour. The following lemma, Lemma 4.1, says that any fair complete multipartite graph with at least three partition classes can be covered with two connected matchings of distinct colours. This is a direct consequence of our Theorem 1.4, and thus there would be no reason to prove such a statement. But, as our aim is to apply the regularity method later, in order to pump up our paths/connected matchings to cycles that cover almost all vertices of the graph, we need a robust version of Lemma 4.1. (This version is given in Lemma 4.2 below.) It will be much easier to extend the proof of Lemma 4.1 for a robust version, than the one of Theorem 1.4. We prefer to give the proof of the exact version first, so that the idea becomes clear to the reader.

15

It will be convenient to formulate the two lemmas only for tripartite graphs. This is justified by Lemma 1.8. Lemma 4.1. Let G be a fair complete tripartite graph on an even number of vertices. If the edges of G are coloured red or blue, then there are two vertexdisjoint connected matchings of distinct colours that together cover all vertices of G. Proof. Say V1 and V2 are the largest two partition classes of G. Let v1 ∈ V1 , v2 ∈ V2 and set G0 = G − {v1 , v2 }. Note that G0 is fair, and unless |V2 | = 1, we may apply induction to obtain two vertex-disjoint connected matchings of distinct colours that together cover all vertices of G0 . On the other hand, if |V2 | = 1, then |V3 | = 1, and thus by fairness, |V1 | = 2. In this case, any two disjoint edges e1 , e2 cover all the vertices of G0 . Clearly, it is either possible to choose e1 , e2 of distinct colours, or in way that they give a monochromatic connected matching. Resuming, in either case there are a red and a blue connected matching MR , MB that cover all of V (G) except v1 , v2 . Let VR be the vertex set of the connected red component of G containing MR , and let VB be the analogue for blue. Even if one of the matchings is empty, note that we can always assume that |VR |, |VB | ≥ 1. Also, VR and VB meet. Indeed, choose any distinct i, j ∈ {1, 2, 3} with VR ∩ Vi 6= ∅ = 6 VB ∩ Vj , say x ∈ VR ∩ Vi and y ∈ VB ∩ Vj . The edge xy is either red or blue, which means that (VR ∩ VB ) ∩ (Vi ∪ Vj ) 6= ∅. In particular, we get VR ∩ VB 6= ∅. Assume v1 v2 is red. Suppose that G cannot be covered with two connected matchings as desired. Thus, {v1 , v2 } ∩ VR = ∅,

(20)

since otherwise we could add v1 v2 to MR . This means that all edges in the cut EG (VR , {v1 , v2 }) are blue.

(21)

In particular, all edges in EG (VR ∩ VB , {v1 , v2 }) are blue. Hence, at least one of v1 , v2 is contained in VB , say v1 ∈ VB . Consequently, V (G) \ {v2 } ⊆ VR ∪ VB . Thus, by (21), we have that V (G) \ (V1 ∪ {v2 }) ⊆ VB .

(22)

Now, if v2 ∈ / VB , then by (20), we have v2 ∈ / VR ∪VB , and hence VR ∩VB ⊆ V2 (since any edge from v2 to VR ∩ VB has some colour). By the argument we used for showing that VR ∩ VB 6= ∅, we know that V1 ∪ V3 is contained either in VR \ VB or in VB \ VR . By (22), this means we have VB \ VR ⊇ V1 ∪ V3 . In either case, we find that VB covers all but at most one partition class. Let MB0 be a largest blue matching in G such that G00 := G − V (MB0 ) is still fair. Since VB covers all but at most one partition class, MB0 is a connected matching. If all edges in G00 are red, then G00 has a red connected perfect matching, and we are done. So assume there is some blue edge uv ∈ E(G00 ). By the choice of MB0 , the graph G00 − {u, v} is not fair. This means neither of u, v lies in the largest partition class V100 of G00 , and, furthermore, |V100 | = |V (G00 )|/2. Thus all edges between V100 and the rest of G00 are red, and hence, we can cover G00 with a red connected matching. 16

We now give a robust version of Lemma 4.1. This is the result we need for applying regularity later. Lemma 4.2. Let G be a fair tripartite graph on n vertices, with partition classes V1 , V2 , V3 , such that |Vi | ≥ 3εn, for some ε with 0 < ε < 1/5. Suppose that d(v) ≥ (1 − ε)(n − |Vi |) for i = 1, 2, 3 and for each v ∈ Vi . If the edges of G are coloured red and blue, then there are two vertex-disjoint connected matchings of distinct colours that together cover all but at most 36εn vertices of G. Proof. Take a red connected matching MR and a blue connected matching MB , which together cover as much as possible of V (G), while leaving G0 := G − V (MR ∪ MB ) fair. Let VR , VB be the vertex sets of the respective colour components, as above we may assume both are non-empty. We assume G0 contains more than 36εn vertices. We claim that for all i, j ∈ {1, 2, 3} with i 6= j and Vi ∩VR 6= ∅ and |Vj ∩VB | > εn, we have (VR ∩ VB ) ∩ (Vi ∪ Vj ) 6= ∅. (23) Indeed, by the degree condition of the lemma, there is an edge from Vi ∩ VR to Vj ∩ VB , and clearly, one of its endpoints lies in VR ∩ VB . This proves (23). Observe that (23) also holds if we interchange the roles of VR and VB . Let Vε := V (G) \ (VR ∪ VB ). Our next aim is to show that there are ` ∈ {1, 2, 3}, X ∈ {R, B} with (a) VX ⊇ V (G) \ (V` ∪ Vε ), and (b) |Vε ∩ Vi | ≤ εn for i ∈ {1, 2, 3}, i 6= `. First of all, since all edges are coloured, note that no vertex in Vε may send an edge to VR ∩ VB . So, by the degree condition of the lemma, we know that |Vε ∩ V` | ≤ εn for all ` with (VR ∩ VB ) \ V` 6= ∅. In particular, if V` is such that |Vε ∩V` | > εn, then VR ∩VB ⊆ V` . This implies |Vε ∩Vi | ≤ εn and |Vε ∩Vj | ≤ εn, where Vi , Vj are the other two partition classes. Moreover, since Vj has at least 3εn vertices, we know that Vj \ Vε has at least εn vertices in either VR \ VB or in VB \ VR , say in VB \ VR . Then by (23), we have that Vi \ Vε is contained in VB \ VR . Again by (23) (interchanging the roles of i and j), we see that also Vj \ Vε is contained in VB \ VR . Thus, either (a) and (b) hold, or |Vε ∩ V` | ≤ εn for all ` ∈ {1, 2, 3}.

(24)

We now assume the latter assertion in order to show that (a) and (b) hold also in this case. Since |V (G0 )| > 36εn, and since G0 is fair, the largest two partition classes 0 V1 , V20 of G0 each contain at least 9εn vertices. For notational ease, we assume Vi0 ⊆ Vi for i = 1, 2. A well-known fact states that any graph H has a subgraph H 0 which has minimum degree at least half the average degree of H. Applying this fact to the bipartite graph spanned by V10 and V20 , in the majority colour, say this is 17

red, we obtain sets U1 ⊆ V10 and U2 ⊆ V20 such that the minimum degree from Ui to U3−i in red is greater than 2εn. In particular, |Ui | > 2εn, for i = 1, 2. By maximality of MR , and since G0 − {u1 , u2 } is still fair for any red edge u1 u2 between U1 and U2 , we know that all edges between U1 ∪ U2 and VR are blue. In particular, the edges between U1 ∪ U2 and some fixed x ∈ VR ∩ VB are blue. Without loss of generality, assume x ∈ / V1 . Since U1 has size greater than 2εn, every vertex y ∈ VR \ V1 sends a blue edge to some blue neighbour of x in U1 . Thus, by the definition of Vε , we have V (G) \ (Vε ∪ V1 ) ⊆ VB , which is as desired for (a). Because of (24), also (b) holds. We have thus shown (a) and (b); let us assume they hold for ` = 1 and X = B. Set Vε0 := Vε \ V1 . Now take a maximal blue matching MB0 in G − Vε0 such that G00 := G − V (MB0 ) is still fair; by (a) we know that MB0 is connected in blue. Assume that |V (G00 )| > 36εn, as otherwise we are done. By maximality of MB0 , for any blue edge in E(G00 −Vε0 ) we have that G00 −e is not fair. Thus, similar as in the proof of Lemma 4.1, either all edges of G00 − Vε0 are red, or G00 contains a spanning balanced bipartite graph H such that any blue edge in H is incident with Vε0 . In either case, we can easily find a red connected matching in G00 covering almost all of G00 as follows. Take a maximal red connected matching MR0 in G00 − Vε0 , or in H − Vε0 , such that the remainder of the graph is still fair. Note that we may assume MR0 6= ∅ as |Vε0 | ≤ 2εn by (b). Let xy ∈ MR0 . The neighbourhoods of x and y in the uncovered part of G00 − Vε0 or H − Vε0 are large enough to span at least one red edge. Also, all edges between x and G00 − Vε0 or H − Vε0 are red, a contradiction to the maximality of MR0 . We now give an analogue of Lemma 4.2 for bipartite graphs. With the obvious exclusion of the proper split colouring, we can cover all 2-edge-coloured balanced bipartite graphs that are almost complete bipartite, with two connected matchings of distinct colours. Lemma 4.3. Let G be a balanced bipartite graph on n vertices, with partition classes V1 , V2 . Suppose G has minimum degree at least (1 − ε)n/2, for some ε with 0 < ε < 1/5. If the edges of G are coloured red and blue, and this colouring is not a split colouring, then there are two vertex-disjoint connected matchings of distinct colours that together cover all but at most 4εn vertices of G. Proof. Take a red connected matching MR and a blue connected matching MB , together covering as much as possible of V (G). Let VR , VB be the vertex sets of the respective colour components. We assume G − V (MR ∪ MB ) contains more than 4εn vertices. As above we see that VR ∩ VB 6= ∅, say there is a vertex x ∈ VR ∩ VB ∩ V2 . Now, if there is an X ∈ {R, B} and i ∈ {1, 2} such that VX covers all but a set Vi0 of at most 2εn vertices of Vi , we can proceed as in the proof of 0 Lemma 4.2. That is, we take a maximal matching MX in G − Vi0 in the colour 0 corresponding to X, and note that MX is connected in this colour. Then all 0 ) must have the other colour, and we can easily cover edges in G − Vi0 − V (MX

18

0 all but at most 4εn vertices of G − V (MX ) with a connected matching in this colour. So, from now on, let us assume that there there is no choice of X ∈ {R, B} and i ∈ {1, 2} as above. That is, we assume

|Vi \ VX | > 2εn for all X ∈ {R, B} and i ∈ {1, 2}.

(25)

Set Vε := V (G) \ (VR ∪ VB ). Since there can be no edges from x to Vε ∩ V1 , we have that |Vε ∩ V1 | ≤ εn. (26) Moreover, if VR ∩ VB ∩ V1 6= ∅, then |Vε ∩ V2 | ≤ εn.

(27)

Since there is no edge between VR \ VB and VB \ VR , there are j ∈ {1, 2} and Z ∈ {R, B} such that |Vj ∩ (VZ \ VY )| ≤ εn, where Y ∈ {R, B} is such that Y 6= Z. In other words, |Vj \ (Vε ∪ VY )| ≤ εn. Together with (25), this implies that |Vε ∩Vj | > εn. So by (26), we have that j = 2, and by (27), we know that VR ∩ VB ∩ V1 = ∅. Hence, V1 \ Vε is covered by the two disjoint sets V1R := V1 ∩ (VR \ VB ) and V1B := V1 ∩ (VB \ VR ). By (25) and (26), we have that |V1R |, |V1B | > εn. Now, as no edge may exist between V1R and VB \ VR , or between V1B and VR \ VB , we see that V2 \ Vε = VR ∩ VB . Observe that thus, all edges between V2 \ Vε and V1R are red, and all edges between V2 \ Vε and V1B are blue. Also, by definition of Vε , all edges between Vε ∩ V2 and V1R are blue, and all edges between Vε ∩ V2 and V1B are red. Additionally, since there are no edges between Vε ∩ V1 and V2 \ Vε = VR ∩ VB , either Vε ∩ V1 is empty, or V2 \ Vε has less than εn vertices. In the first case, we have a split colouring, and in the latter case, we can easily cover all but at most 2εn vertices of G with two monochromatic connected matchings of distinct colours.

5

Covering almost everything with three cycles

5.1

Regularity preliminaries

In this subsection we introduce some very well-known concepts; the reader familiar with regularity is invited to skip this. We start giving the standard definition of regularity. Given ε > 0 and disjoint subsets A, B of the vertex set of a graph G, we say that the pair (A, B) is ε-regular, and of density d(A, B), if, for every pair (A0 , B 0 ) with A0 ⊆ A, |A0 | ≥ ε|A|, B 0 ⊆ B, |B 0 | ≥ ε|B|, we have |d(A0 , B 0 ) − d(A, B)| < ε. When there is no danger of confusion, we simply say that the pair (A, B) is ε-regular. It is well-known and easy to see that together with a pair (A, B), its complement is ε-regular, and of density 1 − d(A, B). For a graph G, we say a partition V0 ∪V1 ∪· · ·∪Vt of its vertex set is ε-regular if the following hold: (i) |V1 | = |V2 | = ... = |Vt | and |V0 | ≤ ε|V (G)|, and 19

(ii) for each i, 1 ≤ i ≤ t, all but at most εt2 of the pairs (Vi , Vj ), 1 ≤ j ≤ t, are ε-regular. We state Szemer´edi’s regularity lemma [20] in its form with a prepartition. We say a partition V1 ∪ V2 ∪ · · · ∪ Vt refines another partition W1 ∪ W2 ∪ · · · ∪ Ws if for each i there is a j such that Vi ⊆ Wj . Theorem 5.1 (Regularity lemma with prepartition). For every ε > 0 and m0 , s ∈ N, there is an m1 ∈ N such that the following holds for each graph G on n ≥ m1 vertices, and with a partition W1 , . . . , Ws of its vertex set. There exists an ε-regular partition V0 ∪ V1 ∪ · · · ∪ Vt such that • V1 ∪ V2 ∪ · · · ∪ Vt refines W1 ∪ W2 ∪ · · · ∪ Ws , and • m0 ≤ t ≤ m1 . It is usual to define a reduced graph of a graph G, for a given ε-regular partition, and a threshold density ρ. This is the graph RG = ([t], E(RG )) which has an edge for each ε-regular pair of density at least ρ. (Note that in spite of the notation RG , the reduced graph RG depends on G, on ρ, and on the given partition.)

5.2

Proof of Theorem 1.6

Theorem 5.1 applied with parameters ε  δ, m0 = 1/ε and s = 2 and s = 3, gives two values for m1 of which we take the larger one. Let G be a fair complete multipartite graph on n > m1 vertices whose edges are coloured red and blue. We assume the colouring is not δ-close to a split colouring. Use Lemma 1.8 to obtain a spanning fair subgraph G0 ⊆ G that is complete tripartite or complete bipartite (possibly G0 = G). In the case that the smallest partition class of G0 has less than δn/20 vertices (and thus G0 is tripartite), we delete this class, and the same number of vertices from the other classes, in a way that the obtained bipartite graph (which we still call G0 ) is balanced. By the proof of Lemma 1.8, we know that |E(G) \ E(G0 )| < δn2 /10. Now, Theorem 5.1 applied to the red subgraph of G0 yields a partition of V (G0 ) refining the bi- or tripartition which is ε-regular in both colours. Using the majority colouring of each pair (that is, we use ρ = 1/2, and in case of a tie we give the edge any colour), we can work with a two-coloured reduced graph RG0 . 0 Note that the non-neighbours of a vertex √ v ∈ V (RG ) correspond to irregular pairs containing v. So, there are at most εt vertices v in RG0 that have more √ than εt non-neighbours in the other partition class(es). We discard these 0 vertices from the reduced √graph RG to ensure that each vertex of the remaining 0 graph RG has at most εt non-neighbours in the other partition class(es). 0 Observe that the size of any class Ci of the bi- or tripartition lies between √ |Cˆi | − |V0 | √ |Cˆi | − εt ≥ − 2 εt |V1 | |V1 |

and

|Cˆi | , |V1 |

(28)

where Cˆi is the class of the bi-/tripartition of G0 corresponding to C √i . So, since G0 is fair, and√since |V0 | ≤ εn, we have that |C1 | ≤ |C2 | + 3 εt, or 0 |C1 | ≤ |C2 | + |C3 | + 2 εt, respectively, if C1 is the largest partition class of RG 0. 20

√ If necessary, we discard at most 3 εt vertices from C1 to make the remaining 00 graph RG 0 fair. 00 0 Resuming, √ we have obtained a bi- or tripartite subgraph RG0 of RG 00on at least (1 − 4 ε)t and at most t vertices. Let the partition classes of RG 0 be 0 0 0 00 0 denoted C , C , and, if applicable, C . In R , every vertex of C has at most 1 2 3 i G0 √ 00 0 εt many non-neighbours outside of Ci0 . Since RG 0 is fair, t − |Ci | ≥ t/2, and 0 so every vertex of Ci has degree at least √ √ t − |Ci0 | − εt ≥ (1 − 2 ε)(t − |Ci0 |), 00 where we sum the degree over both colours. We may √ view RG0 as a√ reduced 00 0 00 graph for a subgraph G of G , with |V (G )| ≥ (1 − 5 ε)n (allowing εn ≥ εn for the exceptional set V0 ). Note that √ |E(G) \ E(G00 )| ≤ δn2 /10 + 5 εn2 ≤ δn2 /5. (29)

Observe that in the case that G0 is bipartite, the colouring in the reduced 00 graph RG 0 might be a split colouring, even if the colouring in G is not a split colouring. But, in this case, note that if we change the colouring of any edge of 00 RG 0 , we no longer have a split colouring. As all our arguments work the same 00 whether we take the threshold for the colouring of E(RG 0 ) to be 1/2 or δ/2, 00 we may assume that we can either change the colouring of RG 0 in a justified 00 way so that the obtained colouring is not a split colouring, or that RG 0 has a 00 00 split colouring, and each red edge of RG corresponds to a pair in G which 0 has blue density < δ/2, and vice versa for the blue edges. So, deleting at most δ|E(G00 )|/2 edges from G00 we can make all pairs monochromatic. Thus, by (29), this means G is δ-close to a split colouring, a contradiction. 00 Therefore, we may assume the colouring in RG 0 is not a split colouring. 00 Since by √ (28), the classes of the bi- or tripartition of RG 0 have size at least 00 δt/20 − 2 εn, we may apply Lemma 4.2 or Lemma 4.3 to RG 0 to obtain two connected matchings, one in each colour, which together cover all but at most √ 72 εt vertices. Now we use a crucial and well-known lemma that has appeared in similar forms before. In its original form it is due to Luczak [15]. The version we use here is close to the one given in from Section 4 of [12]. Lemma 5.2. Let R be the reduced two-coloured graph of a two-coloured graph H, for some γ-regular partition, where each edge of R corresponds to a γ-regular √ pair of density at least γ. If all but at most 72γ|V (R)| vertices of R can be covered with two disjoint connected monochromatic matchings, one of each colour, then H has two disjoint monochromatic cycles, one of each colour, which together cover at least (1 − √ 100 γ)|V (H)| vertices of H. For completeness, let us outline a proof of Lemma 5.2. Sketch of a proof of Lemma 5.2. We first connect in H the pairs corresponding to matching edges with monochromatic paths, following their connections in R. We do this simultaneously for both colours. Note that in total, these paths consume only a constant number of vertices of H. Then we connect the monochromatic paths using the matching edges, blowing up the edges to long paths, where regularity ensures we can use all but a small fraction of the corresponding pairs. This gives the desired cycles. The above argumentation is also explained, rather detailed, in the proof of the main result of [11]. 21

00 Applying Lemma 5.2 √ to the graph RG 0 gives the desired two cycles which cover all but at most 100 εn < δn of our graph G. This finishes the proof of Theorem 1.6.

6 6.1

Covering all vertices with 14 cycles Preliminaries

Call a balanced bipartite subgraph H of an 2n-vertex graph ε-hamiltonian, if any balanced bipartite subgraph of H with at least (1 − ε)n vertices in each partition class is hamiltonian. The next lemma is a straighforward combination of results of Haxell [13] and Peng, R¨odl and Ruci´ nski [17]. A proof of this lemma is given in our companion paper [14]. Lemma 6.1. [14] For any 1 > γ > 0, there is an n0 ∈ N such that any balanced bipartite graph G on at least 2n ≥ 2n0 vertices and of edge density at least γ has an γ/4-hamiltonian subgraph of size at least γ 3024/γ n/3. In order to absorb vertices not covered with the cycles given by Theorem 1.6, we use the following result. Lemma 6.2 (Gy´ arf´ as, Ruszink´o, S´ark¨ozy, and Szemer´edi [11]). There is an n and for any colouring of the n0 ∈ N such that for n ≥ n0 and m ≤ (8r)8(r+1) edges of Kn,m with r colours, there are 2r vertex-disjoint monochromatic cycles that cover the m vertices of the smaller side.

6.2

Proof of Theorem 1.7

Let G be a fair complete k-partite graph on n ≥ n0 vertices, whose edges are 2coloured. We assume n0 to be large enough, its value may be extracted from the proof. By Lemma 1.8, we may assume that 2 ≤ k ≤ 3. Let V1 , V2 , and possibly V3 , be the partition classes of G, where we assume that |V1 | ≥ |V2 | ≥ |V3 |. Set 4 δ := 2−10 . For technical reasons, we split the argument into three cases. The explicit value of δ plays a role only in the last case of the proof. We first discuss the case when |V1 | ≤ |V2 | + |V3 | − δn and |V3 | ≥ δn.

(30)

We pick disjoint subsets U11 , U12 ⊆ V1 , U21 , U23 ⊆ V2 , and U32 , U33 ⊆ V3 of size bδn/4c each. Due to (30), it holds that G − {Uij : i, j ∈ {1, 2, 3}} is fair. Indeed, every graph H with G − {Uij : i, j ∈ {1, 2, 3}} ⊆ H ⊆ G is fair, a fact that we will exploit later. Let G1 := G[U11 , U21 ], G2 := [U12 , U32 ], and G3 := G[U23 , U33 ]. Assuming n0 is large enough, we apply Lemma 6.1 with γ := 1/2 to Gi , i = 1, 2, 3, considering only the edges of the respective majority colour. This gives monochromatic 1/8-hamiltonian (and thus balanced) subgraphs Hi = [W1i , W2i ] of Gi , i = 1, 2, 3 with |W1i | ≥ δn/26053 . (31) S3 Now let H := G − i=1 V (Hi ) and recall that H is fair since G − {Uij : i, j ∈ {1, 2, 3}} ⊆ H ⊆ G. By (30) and by the choice of the sets Uji , each partition 22

class of H has size at least δn/2. Let δ 0 := δ/26148 . Assuming n0 is large enough, we know H is not δ 0 -close to a split colouring, and Theorem 1.6 gives that all but at most δ 0 n vertices of H can be partitioned into two monochromatic cycles. If the set X ⊆ V (G0 ) of uncovered vertices has odd cardinality, we cover one of its vertices with a trivial cycle, and from now on assume |X| is even. Assume |X ∩ V3 | ≥ |X ∩ V1 |, |X ∩ V2 | (the other cases are similar). Partition X into equal-sized sets X1 , X2 with X ∩ Vi ⊆ Xi for i = 1, 2. Assuming n0 is sufficiently large, the choice of δ 0 and (31) give that |X1 | ≤ δ 0 n/2 ≤ |W21 |/1624 .

(32)

So, Lemma 6.2 yields a set C of eight vertex-disjoint monochromatic cycles in [X1 , W21 ] and [X2 , W11 ] that together coverS X1 ∪ X2 . As we may assume that C does not contain any trivial cycles, V ( C) ∩ V (H1 ) is balanced, and each side has at most |W21 |/1624 vertices, S by (32). Since H1 is (1/8)-hamiltonian, the monochromatic graph H1 − V ( C) thus has a hamilton cycle. Also, H2 and H3 each admit a hamilton cycle. In total, we covered V (G) using at most 2 + 1 + 8 + 3 = 14 vertex-disjoint monochromatic cycles. We now discuss the case when k = 2. We use the same method as above. Observe that now, we only need to find one (1/8)-hamiltonian subgraph of G (as opposed to the three graphs H1 , H2 , H3 above). Thus in the last step we will use only one instead of three hamilton cycles. However, we might need three cycles to cover G almost entirely, since the colouring of G might be close to a split-colouring (a covering with 3 cycles does exist by the remark after Theorem 1.6). Since we may assume that none of these three cycles is trivial, we may assume the set X of vertices not covered with the three cycles to have even cardinality. In this way we avoid another cycle. This means we can cover G with 3 + 8 + 1 = 12 vertex-disjoint monochromatic cycles in total. To complete the proof of Theorem 1.7, we now consider the case when k = 3, but (30) is violated. Essentially, we proceed as in the case of k = 2. If n is odd, we pick a single vertex v from V1 as a cycle. Since G − v is fair, we may simply assume that n is even. No matter whether we have |V3 | < δn, or we have |V3 | ≥ δn and |V1 | > |V2 | + |V3 | − δn, we proceed as follows. Delete a set W of |V2 | + |V3 | − |V1 | < δn vertices from V3 . Consider the bipartite graph G0 spanned by V1 and the remains of V2 ∪ V3 , which we call V20 . Note that G0 is complete bipartite and balanced. Let U1 ⊆ V1 and U2 ⊆ V2 have size dn/4e each. (Note that clearly |V2 | ≥ n/4.) Assuming n0 is large enough, we may apply Lemma 6.1 with γ = 1/2 to the graph induced by the majority colour to find a (1/8)-hamiltonian subgraph H = [U10 , U20 ] of G0 [U1 ∪ U2 ], with |U10 | = |U20 | ≥ n/26052 . Assuming n0 is large enough, the remark after Theorem 1.6 gives three monochromatic vertex-disjoint cycles that cover all but a set W 0 of at most δn many vertices of G0 − V (H). We may assume that none of these cycles is trivial, and thus |W 0 ∩ V1 | = |W 0 ∩ V20 |. Partition W ∪ W 0 into two equal-sized sets W1 and W2 such that for i = 1, 2, all edges between Wi and Ui0 are present. As 1624 |W1 | < δn/297 ≤ n/26052 ≤ |U20 |, we may apply Lemma 6.2 to cover each of W1 , W2 with four vertex-disjoint monochromatic cycles, which we again assume to be non-trivial. Since H is 23

(1/8)-hamiltonian, the yet unused vertices of H span a hamiltonian cycle. In total, we covered G with 1+3+8+1 = 13 vertex-disjoint monochromatic cycles.

7

Conclusion

Our Theorem 1.4 and the earlier result of Pokrovskiy together cover all cases of monochromatic path covers in 2-edge-coloured multipartite graphs. It would be natural to investigate the same problem for more colours, in the spirit of Conjecture 1.1. For bipartite graphs whose edges are coloured with r colours, Pokrovskiy [18] conjectures that there is a partition into 2r − 1 monochromatic paths, and he shows this number is best possible. For k-partite graphs with k ≥ 3 a partition into less monochromatic paths, perhaps r paths, might be possible, in analogy to the case r = 2 we treated here. For cycle partitions, as said in the introduction, we believe that Theorem 1.6 is not best possible. However, recalling the example given in the introduction, a split colouring between a partition class of size n/2 and the rest of the graph, it is not always possible to cover all vertices of any large fair multipartite graph with two vertex-disjoint monochromatic cycles. This remains true even if the colouring of this graph is far from a split colouring (the other classes might have size n/4 each). Even if a partition class containing half of the vertices of the graph is forbidden, such constructions are possible. Indeed, given a properly split-coloured balanced complete bipartite graph one can add a third partition class consisting of a single vertex v3 seeing only one colour. Then the obtained graph cannot be partitioned into two monochromatic cycles. One can even add a fourth partition class consisting of a single vertex v4 seeing only the other colour, while giving v3 v4 any colour. Still the obtained graph has no partition into two monochromatic cycles. Problem 7.1. For k ≥ 3, under which conditions does a fair complete k-partite graph with a 2-colouring of the edges admit a partition into two monochromatic cycles? One candidate for a sufficient condition in Problem 7.1 could be balancedness, that is, having partition classes of equal size. In any case, we think that considering balanced multipartite graphs is a reasonable restriction that might be of its own interest. The next natural step is to extend Theorems 1.3 and 1.6 to more colours. We have seen here that three disjoint monochromatic cycles can cover all but o(n) of the vertices of any large enough multipartite graph, whose edges are two-coloured (where three cycles are only needed if the colouring is very close to a split colouring). In [14], together with Richard Lang, we prove that in large enough 3-edge coloured balanced complete bipartite graphs, five disjoint monochromatic cycles suffice to cover the graph almost entirely. In analogy to Pokrovskiy’s conjecture for path covers [18] mentioned above, it might always be possible to cover almost all vertices of any large enough multipartite graph, whose edges are r-coloured, with 2r − 1 disjoint monochromatic cycles. Perhaps the number of cycles needed can even be dropped to r in kpartite graphs with k ≥ 3. Maybe this is even possible in bipartite graphs with a colouring sufficiently far from a specific problematic colouring. 24

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